Problem 78 - Chemistry and Chemical Reactivity

Question

A tree is exactly \(10 \mathrm{m}\) tall. (a) What must be the total molarity of the solutes if sap rises to the top of the tree by osmotic pressure at \(20^{\circ} \mathrm{C} ?\) Assume the groundwater outside the tree is pure water and that the density of the sap is \(1.0 \mathrm{g} / \mathrm{mL} .\left(1 \mathrm{mm} \mathrm{Hg}=13.6 \mathrm{mm} \mathrm{H}_{2} \mathrm{O} .\right)\) (b) If the only solute in the sap is sucrose, \(\mathbf{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) what is its percent by mass?

Step-by-Step Solution

Answer

(a) Total molarity of solutes is approximately 0.04015 mol/L. (b) Percent by mass of sucrose is approximately 1.374%.

Step 1: Understanding the Problem

We need to find the molarity of the solutes that can create an osmotic pressure sufficient to raise sap 10 meters in a tree. We'll also find the mass percent of sucrose in the sap at 20°C.

Step 2: Convert Height to Pressure

Convert the height of the water column (10 m of sap) into pressure using the density of sap (1.0 g/mL or 1000 kg/m³). Remember that 1 m of H₂O corresponds to 9,806.65 Pa. Therefore, 10 m of H₂O corresponds to \[ P = 10 \times 9,806.65 = 98,066.5 \, \text{Pa}. \]

Step 3: Convert Pressure from Pascal to mmHg

Use the conversion factor between mmHg and mmH₂O to find the equivalent osmotic pressure in mmHg.\[ 1 \, \text{mmHg} = 13.6 \, \text{mmH}_2\text{O}, \] and using \[ P_{mmHg} = \frac{98,066.5}{133.322} \approx 735.5 \, \text{mmHg}. \]

Step 4: Use Osmotic Pressure Equation

Using the osmotic pressure equation \[ \Pi = iMRT, \] where \( \Pi = 735.5 \, \text{mmHg} = 735.5/760 \times 101325 = 98,066.5 \, \text{Pa}, \) \(i\) is the van 't Hoff factor (assuming 1 for non-dissociating sucrose), \( R = 8.314 \, \text{J/mol K}, \) and \( T = 293 \, K (20°C), \) solve for \( M \):\[ M = \frac{98,066.5}{8.314 \times 293} \approx 40.15 \, \text{mol/m}^3. \]

Step 5: Convert Molarity to Percent by Mass

The molarity of sucrose is approximately 0.04015 mol/L. Convert this to percent by mass:\[ \text{Molar mass of sucrose} = 12 \times 12 + 22 \times 1 + 11 \times 16 = 342.296 \, \text{g/mol}. \]Thus, the concentration by mass is\[ c = 0.04015 \, \text{mol/L} \times 342.296 \, \text{g/mol} = 13.74\, \text{g/L} \approx 1.374\, \% \text{(w/v)}. \]

Key Concepts

Molarity

Molarity is a way to express the concentration of a solution. It tells us how many moles of a solute are present per liter of solution. In this exercise, we need to calculate the molarity that would create enough osmotic pressure to move sap up a tree of 10 meters at 20°C.
Molarity is essential because it directly impacts the solution's properties, like osmotic pressure. The formula to find molarity (\( M \)) from osmotic pressure is:

• \( \Pi = iMRT \)
• \( \Pi \) is the osmotic pressure
• \( i \) is the van 't Hoff factor (1 for sucrose as it doesn't dissociate)
• \( R \) is the gas constant \( 8.314 \, \text{J/mol K} \)
• \( T \) is the temperature in Kelvin

By rearranging this formula, we find the molarity of sucrose in sap necessary for the osmotic process.

Percent by Mass

Percent by mass, also known as mass percent, is used to describe the concentration of a particular solute in a solution. It represents the mass of the solute divided by the total mass of the solution, multiplied by 100 to express it as a percentage.
When discussing the sucrose concentration in sap, we convert the molarity to a percent by mass. To do this:

• Find the molar mass of sucrose: \( \text{C}_{12} \text{H}_{22} \text{O}_{11} \) = 342.296 g/mol
• Use the molarity to convert to grams per liter
• Calculate the percentage: multiply the mass of sucrose by 100 to obtain the mass percent

This process allows us to understand how much sucrose is present in relation to the entire solution, an important consideration in both chemistry and biology.

Colligative Properties

Colligative properties are the physical properties of solutions that depend on the number of solute particles in a solvent, not the identity of those particles. Osmotic pressure, a key concept in this exercise, is one such property.
As the sap rises in the tree due to osmotic pressure, it illustrates how the presence of solute particles affects overall pressure and concentration. Other colligative properties include:

• Boiling point elevation
• Freezing point depression
• Vapor pressure lowering

Understanding these properties helps explain why a solution behaves differently than pure solvent. In trees, osmotic pressure enables the movement of sap, highlighting the vital role of these principles in natural processes.